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HAWKES LEARNING SYSTEMS math courseware specialists

Copyright © 2010 Hawkes Learning Systems.

All rights reserved.

Hawkes Learning Systems:

Intermediate Algebra

Section 7.1a: Quadratic Equations: The Square

Root Method




Objectives o Solve quadratic equations by factoring o Solve quadratic equations by using the definition of a square root.




Solving Quadratic Equations by Factoring

When the solutions of a polynomial equation can be found by factoring, the method depends on the zero-factor property .

Zero-Factor Property

If the product of two factors is 0, then one or both of the factors must be 0. For factors a and b,

If

 

0, then a



0 or b



0 or both.

For example,

 x



1

 x



2

 

0 x 1 0 x 2 0




Solving Quadratic Equations by Factoring

An equation that can be written in the form ax

2

 bx

 

0

Where a , b , and c are real numbers and 0 is called a quadratic equation .




To Solve an Equation by Factoring

1. Add or subtract terms so that one side of the

equation is 0.

2. Factor the polynomial expression.

3. Set each factor equal to 0 and solve each of the resulting equations.

Note: If two of the factors are the same, then the d solution is said to be a double root or a root d of multiplicity two.

HAWKES LEARNING SYSTEMS



math courseware specialists

Example 1: Solving Quadratic Equations by

Factoring

Solve the following quadratic equation by factoring. x

2



3 x



18 x

2



3 x



18



0

Subtract 18 from both sides.

One side must be zero.

 x



3

 x



6

 

0 Factor the left-hand side. x 3 0 or x

 

3 x 6 0 x



6

Set each factor equal to 0.

Solve each linear equation.





Example 2: Solving Quadratic Equations by

Factoring

Solve the following quadratic equation by factoring.

2 x

2



16 x

 

32

2 x

2



16 x



32



0

2

 x

2 

8 x



16





0

2

 x



4



2 

0 x 4 0 x



4

The solution is a double root.




Solving Quadratic Equations by Factoring o Quadratic equations may have nonreal complex solutions. o In particular, the sum of two squares can be factored as the product of complex conjugates.

o For example, x

2 

81

  x



9 i

 x



9 i



.


Copyright © 2010 Hawkes Learning Systems. math courseware specialists All rights reserved.

Example 3: Quadratic Equations Involving the

Sum of Two Squares

Solve the following equation by factoring.

x

2 

49



0

 x



7 i

 x



7 i

 

0 x

 

0 x

 

7 i x

 

0 x



7 i

Check:

 

2 

4 9



0

49 i

2 

49



0

49 49



0

0



0

 

2 

4 9



0

49 i

2 

49



0

49 49



0

0



0


Square Root Property



v x

2

 c

 x

 a



2  c x

  c .

x a x a c

Note: If c is negative, then the solutions will be nonreal.





Using the Definition of Square Root and the


Consider the equation x

2



8

Allowing that the variable, x , might be positive or negative, we use the definition of square root, x

2

 x .

Taking the square root of both sides of the equation gives: x



8

So we have two solutions, x



8 and x

 

8, or x

 

8 .





Using the Definition of Square Root and the


Similarly, for the equation

 x



4



2 

3 the definition of square root gives x 4 3 .

This leads to the two equations and the two solutions, as follows: x 4 3 x 4 3 x 4 3 x 4 3 x 4 3




Example 4: Using the Square Root Property

Solve the following quadratic equations by using the

Square Root Property. a.

 x



5



2 

3 x 5 3 b. x 5 3 y

2  

16 y

  

16 y

 

4 i






Section 7.1b: Quadratic Equations:

Completing the Square




Objectives o Solve quadratic equations by completing the square o Find polynomials with given roots.




Completing the Square o Recall that a perfect square trinomial is the result of squaring a binomial. o Our objective here is to find the third term of a perfect square trinomial when the first two terms are given. This is called completing the square .




Completing the Square

Perfect Square

Trinomials x

2 

12 x



36 x

2 

8 x



16 x

2 

2 ax

 a

2 x

2 

2 ax

 a

2

Equal Factors

 x



6

 x



6

 x



4

 x



4



 x

 a

 x

 a



 x

 a

 x

 a





Square of a

Binomial



 x



6



2

 x



4



2

 x

 a



2 x

 a



2




Completing the Square ax

2  bx

  c .

a



1 , so that the coefficient of x

2

1 x

 b x

  a c a

.

x 2 result, and add this to both sides.

Step 4: The trinomial on the left side will now be a perfect square. That is, it can be written as the square of an algebraic expression.




Example 1: Completing the Square

Add the constant that will complete the square for the expression. Then write the new expression as the square of a binomial. x

2 

18 x

Solution: Since the leading coefficient is 1, we can begin to complete the square. We take half of the coefficient of the x term and square the result.

1



9

2

 

2 

81





Adding 81 x



18 x square trinomial x

2 

18 x



81

This can be factored as

 x



9



2

.

Thus completing the square gives us: x

2 

18 x



8 1

  x

 

2

9 .





Solve the following quadratic equation by completing the square. x

2  

6 x



10

2

1    

3

Take half of the coefficient of the x term.

 

2 

9 Square the result.




Example 2: Completing the Square (Cont.)

Adding 9 to both sides of the original equation will result in a perfect square trinomial. x

2  x 6

2 x

 

6 9



10



9

Add 9 to both sides

(completing the square).

 x



3



2 

19 Factor the left-hand side.

Use the Square Root Property to solve. x 3 x 3

19

19 or or x 3 x 3

19

19

We write these two solutions as: x 3 19





Solve the following quadratic equation by completing the square.

2 x

2 

8 x



24



0 x

2 

4 x



12



0

Divide each term by 2 so that the leading coefficient will be 1. x

2 

4 x



12 Isolate the constant term.

2

1  



 

2 

4

Take half of the coefficient of the x term, and square it.




Example 3: Completing the Square (Cont.)

Add 4 to both sides of x 2 

4 x



1 2 .

x

2 

4 x



4



12



4

 x



2



2 

16 x 2 16

Factor the left-hand side of the equation.

Use the Square Root

Property. x 2 4 or x 2 4 x



2 or x

 

6




Writing Equations with Known Roots

To find the quadratic equation that has the given x 4 i each equation. x i 0 x i 0

Set the product of the two factors equal to 0 and simplify.

 x 4 i

 x 4 i



0

Continued on the next slide…




Writing Equations with Known Roots

Regroup the terms to present the product of complex conjugates. This makes the multiplication easier.

 x 4 i

 x 4 i



0

 x

 i  

 x

 i

 x



4



2 i

2

0

0 x

2 

8 x

   i

2  

1.

x

2 

8 x



17



0




Example 4: Equations with Known Roots

Find the quadratic equation that has the given roots. x 7 5 and x 7 5 x 7 5



0 x 7 5



0

Get 0 on one side of each equation. x 7 5 x 7 5 



0

Set the product of the two factors equal to 0.

 x 7



5

 x 7



5 



0





Example 4: Equations with Known Roots

(Cont.)

Simplify and solve.

 x 7



5

 x 7



5 



0

 x



7



2 

 

2

5



0 x

2 

14 x

  

0 x

2 

14 x



44



0




Hawkes Learning Systems


Section 7.2: Quadratic Equations: The Quadratic

Formula




Objectives o Write quadratic equations in standard form.

o Identify the coefficients of quadratic equations in standard form.

o Solve quadratic equations using the quadratic formula.

o Determine the nature of the solutions (one real, two real, or two non-real) for quadratic equations using the discriminant.




Standard Form of the Quadratic Equation

The standard form of the quadratic equation is ax

2  bx

 

0

We are interested in developing a formula that can solve quadratic equations of any form. This formula

will always work, though other techniques may be more convenient to use.

We want to solve the standard form equation of the quadratic formula for x in terms of a , b and c .




Development of the Quadratic Formula ax 2  bx

 

0 ax

2  bx

  c

Standard form.

Add –c to both sides.

x

2  b x

  a c a x

2  b a x



4 b a

2

2

   a 4 b a

2

2 x

 b

2 a

2 c b a 4 a

2

2

Divide both sides by a so that the leading coefficient is 1.

Complete the square.

Simplify.

Continued on the next page…




Development of the Quadratic Formula x

 b

2 a

2

 

4 ac

 b

4 a 2 4 a

2

2 x x

 b

2 a

2

 b

2 

4 ac

4 a 2



2 b a

  b

2 

4 ac

2 a x



  b

2 

4 ac

2 a

Find LCM of the denominators on the right side.

Combine the fractions on the right side of the equation.

Square Root method.

The quadratic formula.




Development of the Quadratic Formula

The quadratic formula ,

  b

2 

4 ac x



2 a

, can ALWAYS solve quadratic equations of any form.

Because it is so useful, you should memorize the quadratic formula.




Discriminant b



4 ac discriminant . The

discriminant determines the number of solutions to the given quadratic equation.

If the discriminant is:

Discriminant

Positive b

2 

4 ac



0

Zero b

2 

4 ac



0

Negative b

2 

4 ac



0

Number of

Solutions

Solution(s) Real or Non-real

2

Real

1

Real

2

Non-real


Copyright © 2010 Hawkes Learning Systems. math courseware specialists

Solve Quadratic Equations Using the


Quadratic Formula

Ex: Solve the quadratic equation using the quadratic formula.

1 x

2 a





1

2

, x b =

1

2 , c

0

= 1

Compare it to the standard quadratic ax

2   

0 equation to find a, b and c.

 

2 

4

  

4 4 0

Solve the discriminant by plugging a, b and c into . Since the discriminant is 0, there is one real solution.

x



  

2



  x





2

2

 

1

0

Use the quadratic formula x



  b

2 

4 ac

2 a to find the solution, x.

We know that the discriminant is 0, so we can just plug that in.

Solution: ̶ 1



Example 1: Solve Quadratic Equation Using the Quadratic Formula

Solve the quadratic equation using the quadratic formula.

2 x x a

 

1 , b = 1 , c =



1


2   

0 equation to find a, b and

c.

Discriminant:

 

2 

4

   Solve the discriminant by plugging a, b and c into .

 

3

Number of solutions:

Real or non-real: non



2 real

What does it mean for a discriminant to be negative?


x

 x





Example 1: Solve Quadratic Equation Using the Quadratic Formula (Cont.)

Use the quadratic formula

  

2

 

 

3 x



  b

2 

4 ac

2 a to find the solution,

x. We know that the discriminant is ̶ 3, so we can plug that in.

1



2

3



1

 

2

3 x



1

 i 3

2

Solutions: x



1

 i 3

2 and x



1

 i 3

2



Example 2: Solve Quadratic Equation Using the Quadratic Formula

Solve the quadratic equation using the quadratic formula.

x

2

3 x 2 0 a



  

1 , b



3 , c



2


2   

0 equation to find a, b and

c.

Discriminant:

 

2 

4

   Solve the discriminant by plugging a, b and c into .



1

Number of solutions:

Real or non-real: real

2

What does it mean for a discriminant to be positive?


x

 x





Example 2: Solve Quadratic Equation Using

  

2



  the Quadratic Formula (Cont.)

Use the quadratic formula

1 x



  b

2 

4 ac

2 a to find the solution,

x. We know that the discriminant is 1, so we can plug that in.

2

Solutions: x

 

1 and x

 

2






Section 7.3: Applications- Quadratic Equations




Objective

Solve applied problems by using quadratic equations.




Strategy for Solving Word Problems

1. Read the problem carefully.

2. Decide what is asked for and assign a variable to the unknown quantity.

3. Draw a diagram or set up a chart whenever possible.

4. Form an equation (or inequality) that relates the information provided.

5. Solve the equation or inequality.

6. Check your solution with the wording of the problem to be sure that it makes sense.




The Pythagorean Theorem o Problems involving right triangles often require the use of quadratic equations.

o In a right triangle, one of the angles is a right angle

(measures ), and the side opposite this (the longest side) is called the hypotenuse .

o The other two sides are called legs . hypotenuse leg leg




The Pythagorean Theorem

In a right triangle, the square of the hypotenuse c

2  a

2  b

2 b a c




Example 1: The Pythagorean Theorem

The width of a rectangle is 5 yards less than its length. If one diagonal measures 25 yards, what are the dimensions of the rectangle?

Solution: Draw a diagram for problems involving geometric figures whenever possible.

Let x



length of the rectangle x 5 width of the rectangle

Then, by the Pythagorean Theorem, x

 x

2   x



5

  

2

5

25 x




Example 1: The Pythagorean Theorem (Cont.)

Now, solve the quadratic equation. x

2   x



5

  

2 x

2  x

2 

10 x



25



625

2 x

2 

10 x

   

Length:

Width:

2



2 x

2 

10 x



600



0 x

2



5 x



300





0

2

 x



20

 x



15

 

0 x



20 x

 

15 x 5 15

A negative number does not fit the conditions of the problem.




Projectiles relates to the height of a projectile such as a thrown ball, a bullet or a rocket. h

 t

 v



0 height of object, in feet.

time object is in the air, in seconds. beginning velocity, in feet per second. h



0 h



0

0 beginning height

. if object is initially at ground level.


Example 2: Projectiles



A bullet is fired straight up from 6 feet above ground level with a muzzle velocity of 420 ft per sec. When will the bullet hit the ground?

Solution: v

0



420 ft per sec. h



0

6

The bullet hits the ground when h



0 .

h

 

16 t 2  v t

0

 h

0

0

 

16 t

2 

420 t



6

0

 

8 t

2 

210 t



3

Divide both sides by 2

.


Example 2: Projectiles



0

 

8 t

2 

210 t



3

Use the quadratic formula to solve the equation. t



  b

2 

4 ac

2 a t





210

 

210

2



2

 

   t

 t





210



44196



16

26.26

t

 

0.01

Therefore, the bullet hits the ground in

26.26

seconds.




Example 3: Cost per Person

The Ski Club is planning to charter a bus to a ski resort. The cost will be $200 and each member will share the cost equally. At the last minute, 30 more member decide to go on the trip. The cost to each of the members will be $6 less. How many members are going to the ski resort now?

Let x

 the original number of club members going on the trip.

x



30

 the actual number of club members that took the trip.

Initial cost per member



Final cost per member



Difference in cost per member

200 x



200 x



30



6




Example 3: Cost per Person (Cont.)

Multiplying each term by the LCM of the denominators results in the following:

200

  

30

  x

200 x



30

  

30



6

 

30



200

 x



30

 

200 x



6 x x



30



200 x

  x



6

 

30



6000



6 x 2 

180 x

0



6 x

2 

180 x



6000

0

 x 2 

30 x



1000

Factor out 6 .




Example 3: Cost per Person (Cont.) x

2 

30 x



1000



0

Use the quadratic formula to solve. x



  b

2 

4 ac

2 a x



   

2 

 

  

1000

 x



2 x



20 x

 

50

The number of members that actually went to the ski resort is

20



30



5 0 .


Example 4: Geometry



The Smiths have a rectangular swimming pool that is 5 ft longer than it is wide. The pool is completely surrounded by a concrete deck that is 3 ft wide. The total area of the pool and the deck is

1750 ft . Find the dimensions of the pool.

Let w 6 width of the pool and the deck w



11

 length of the pool and the deck

 w



6

 w



11

 

1750

3 ft w

3 ft w



5

3 ft

3 ft




Example 4: Geometry (Cont.)

 w



6

 w



11

 

1750

 w

2 

17 w



1684



0

Use the quadratic equation to solve. w



  b

2 

4 ac

2 a w



 

17

2 

4 1



1684

 

  

Therefore, the length of the pool is 33.41 feet , and the width of the pool is

38.41 feet . w



17 w



33.41

2

7025 or w 5 38.41

w

 

50.41






Section 7.4: Equations in Quadratic Form




Objectives o Make substitutions that allow equations to be written in quadratic form.

o Solve equations that can be written in quadratic form.

o Solve equations that contain rational expressions.




Solving Equations in Quadratic Form ax

 bx c 0 a



0 .

The equations

2 1 x

4 

7 x

2 

12



0 are not quadratic equations, but they are in quadratic

form because the degree of the middle term is one-

half the degree of the first term. Specifically,

1

2

  

2 and

1

2

2

 



1

3

Degree of the first term

Degree of the second term

Degree of the first term

Degree of the second term




Solving Equations in Quadratic Form

Solving Equations in Quadratic Form by Substitution

1. Look at the middle term.

2. Substitute a first-degree variable, such as u , for the variable expression in the middle term.

3. Substitute the square of this variable, u² , for the variable expression in the first term.

4. Solve the resulting quadratic equation for u .

5. Substitute the results back for u in the beginning substitution and solve for the original variable.




Example 1: Solving Equations in Quadratic Form

Solve the equation in quadratic form by substitution.

x

6 

7 x

3 

12



0 Step 1: Look at the middle term.

u

2  u

 x

3

7 u



12



0

Step 2: Substitute a first-degree variable, such as u, for the variable expression in the middle term.

Step 3: Substitute the square of this variable, u², for the variable expression in the first term.

 u



3

 u



4

 

0 u



3 or u



4

Step 4: Solve the resulting quadratic equation for u.

x

3 

3 or x

3 

4

Step 5: Substitute the results back for u in the beginning substitution and solve for the original variable.

x



3 3 or x



3 4






5 x

5 

9 x 2



18

5




Step 2: Substitute a first-degree variable, such as u, u

 x 2 for the variable expression in the middle term.

u

2 

9 u



18



0


 u



6

 u 3



0 Step 4: Solve the resulting quadratic equation for u.

u



6 or u



3 x 2

5



5

6 or x 2



3


x

5 

36 or x

5 

9 x



5 36 or x



5 9




Example 3: Solving Equations in Quadratic Form x


x

2

3



3 x

1

3

 


u

2  u

 x

1 3

3 u

 

0



 u



1

 u



2

 

0 u

 

1 or u

 

2 x

1 x

3

 

1 or x

 

3 x

1

3

 

2

 

3

 

1 or x

 

8








x



2 

10 x



1 

16




u

2 u

 x



1



10 u



16



0



u

 u



2

 u 8



0



2 or u



8


x



1 

2 or x



1 

8


x



1

2

or x



1

8








x



6



2 

4

 x




Step 2: Substitute a first-degree u x 6 variable, such as u, for the variable expression in the middle term. Step 3: u

2 

4 u

 

0

Substitute the square of this variable,

u², for the variable expression in the first term.

 u



2

 u



2

 

0 u



2


 x



6

 

2 x

 

4





Solving Equations with Rational Expressions

Recall that to solve equations with rational expressions, first multiply every term on both sides of the equation by the least common multiple (LCM) of the denominators. This will “clear” the equation of fractions. Remember to check the restrictions on the variables. That is, no denominator can have a value of 0.



Example 6: Solving Equations with Rational

Expressions

Solve the following equation containing rational expressions.

4 x

3



2

 x

2



3

 x x



3

This equation is not in the quadratic form. Multiply both sides of the equation by x



1

2

, 3



4 x



2

 x



3





4 x

3



2

 x

2



3



 

4 x



2

 x



3



 x x



3



3

 x

  x

Simplify.



2

 

4 x



2



3 x

  x

 

0



4

4 x x

2

2





2

13 x x



5

Continued on next slide...




Expressions (Cont.)

0



4 x

2 

13 x



5 Use the quadratic formula to solve.

x



13

  

13



2 

 

   x



13

 

8 x



13



249

8

There are two solutions to this equation.




Expressions

Solve the following equation containing rational expressions.

2 x

1



3

 x

2



2

 x x



2

This equation is not in the quadratic form. Multiply both sides of the equation by x





2

3

, 2



2 x



3

 x



2





2 x

1



3

 x

2



2



 

2 x



3

 x



2



 x x



2



1

 x

  x

Simplify.

 

2 x



3

 x

  x

 

0



2

2 x x

2

2





3

2 x x



4

Continued on next slide...




Expressions (Cont.)

0



2 x

2 

2 x



4 Use the quadratic formula to solve.

x



2

  

2 

 

   x



2



4 32



2



36

4 4 x



4

 

1 or 2 x





3

, 2

2

, so the only solution is x

 

1




Example 8: Solving Higher-Degree Equations

Solve the higher degree equation.

x

5 

9 x



0



4 

9





0



2 

3

 x 2 3



0

Factor out x.

Factor the difference of two squares.

x



0 or x

2  

3 or x

2 

3 x



0 or x

  i 3 or x

 

3

Solve for x.

x



0, i 3,

 i 3, 3,



3 There are five solutions for x.




Example 9: Solving Higher-Degree Equations

Solve the higher degree equation.

x

3

8 0

 x



2

 

 x



2

  x 2 

2 x



4





0 x

  i 3

 x

  i 3





0

Factor out (x – 2).

Factor the quadratic equation.

x



2 or x

   i 3 or x

   i 3

Solve for x.

x

   i

  i 3 There are three solutions for x.






Section 7.5: Graphing Parabolas




Objectives o Graph a parabola (a quadratic function) and determine its vertex, domain , range, line of symmetry and zeros.

o Solve applied problems by using quadratic functions and the concepts of maximum and minimum.




Introduction to Quadratic Functions

Concepts related to various types of functions and their graphs include domain, range and zeros.

In this section we will include a detailed analysis of

quadratic functions, functions that are represented by quadratic expressions.




Introduction to Quadratic Functions of the graph can be investigated by plotting several points.

x



1

0 x

2 

4 x



3

 

2

 

3

 

2    

3 y

8

3

1

2

1

2

3

7

2

4

5

 

 

2



4

 

 



 

 

2   

 



3

2

2 

4 2



3

3

 

2    

3

 

 

2



4

 

 

 

 

2

2





 

 







3

3

3



1

3

8

5

4

5

4

0

0





This curve is called a parabola .

  point” of the parabola and is called the vertex . The line is the line of symmetry or axis of symmetry for the parabola.

That is, the curve is a “mirror image” of itself with respect to the line .





Quadratic Functions

A quadratic function is any function that can be written

   ax

2  bx

 c where a , b , and c a



The graph of every quadratic function is a parabola.

Parabolas that open up or down are vertical parabolas and parabolas that open left or right are horizontal parabolas . Horizontal parabolas do not represent functions.





We will discuss quadratic functions in each of the following five forms where a , b , c , h and k are constants: y

 ax

2 y

 ax 2  k y

   h



2 y

   h



2  k y

 ax

2  bx

 c


Quadratic Functions



Functions of the Form y

 ax

2

If a



0 If a



0

The graph lies Below the x-axis Above the x-axis

Vertex

 

This is the only point where the graph touches the x-axis.

Domain

Range y



0

(All real numbers) y



0


Quadratic Functions



Domain: {x|x is any real number}

Range: {y|y ≥ 0}

In interval notation:

  

0,

 


Range: {y|y ≤ 0}


   

,0




Quadratic Functions



The graphs on the previous slide illustrate the following characteristics of quadratic equations of the form : a. If , the parabola “opens upward.” b. If , the parabola “opens downward.”


Quadratic Functions




 ax

2  k

The y-value

Shift y

 ax

2

Vertex

Line of

Symmetry

If k



0

Decreases by k

Down k units x



0

If k



0

Increases by k


Quadratic Functions




Range: {y|y ≥ k}


   k ,

 


Range: {y|y ≤ k}


   

, k




Quadratic Functions


   h



2



If a



0

The graph lies Below or on the

x-axis

Vertex

The graph opens

Down

If a



0

Above or on the

x-axis

Up


Quadratic Functions




Range: {y|y ≥ 0}


  

0,

 


Range: {y|y ≤ 0}


   

,0




Quadratic Functions



y

  

2

If h



0

Decreases by h

If h



0

Increases by h The x-value

Shift y

 ax

2

Line of

Symmetry x

 h


Quadratic Functions



In summary,

Graph Shift of y

 ax

2 y

 ax

2 y

 ax

2  k y

   h



2

Does not apply

Up or Down

Left or Right

Vertex

Line of

Symmetry x



0 x



0 x

 h


Quadratic Functions



line of symmetry and the vertex, and state the domain and range of each function.

Solution: has x = 0 for its line of symmetry.

The parabola opens upward.

Vertex:

 

a = 2, so a is positive.

k

 

3

 





Domain: or





  y y

  

Quadratic Functions



True of all functions of the form .

 




Example 1: Quadratic Functions

2 y

 

 x



2  of symmetry and the vertex, and state the domain and range of each function.

Solution: x



5

Line of symmetry is .

2

The parabola opens .

Vertex: 

5

2

,0  downward

 

2 has x = h for its line of symmetry.

a = ̶ 1, so a is negative.

h



5

2

5

, so the vertex is .

 2 





Domain: or





  y y



0



Quadratic Functions



True of all functions of the

 

2

 

,0



( )


Quadratic Functions y

   

2  k



 

 

2

For example, the graph of the function y

 

2 x 5 y units and to the

5 right 3 3,5 y

 x

2 y



 x



1

2

 2



1

2

2

2 units.

   

 


Quadratic Functions




Quadratic Functions



To easily find the vertex, line of symmetry, range, and to graph the parabola, we want to change the general y

  

2  k

This can be accomplished by “completing the square” using the following technique.

   ax

2  bx

 c



 a x

2  b a x



 c

 a



 x

2  b a x

 b a

2

2



4 4 b a

2

2

Write the function.

Factor a from the first two terms.





 c

Complete the square.



Quadratic Functions



 a



 x

2







 b b x a



4 a

2 b

2



2







4



2 a 4 a b 2

4 a

2

 c

Move the negative factor out of the parentheses, first multiplying by a.

Write the square of the binomial and simplify.

In terms of the coefficients a, b and c, x

  b

2 a is the line of symmetry





 b

2 a

4

4

 a

2





Note: the vertex must lie at





 b

2 a

, f



 b

2 a 





.


Quadratic Functions



Note!

Rather than memorize the formulas for the coordinates of the vertex, you should just remember that the x

 

2 b a value for x in the function will give the y-value for the vertex.




Zeros of a Quadratic Equation

The points where a parabola crosses the x-axis, if any, are the x-intercepts. This is where . These points are also called the zeros of the function . We find these points by substituting 0 for y and solving the resulting quadratic equation.

y

 ax 2  b x

 c

0

 ax

2  bx

 c quadratic function quadratic equation

If the solutions are non-real complex numbers, then the graph does not cross the x-axis.




Zeros of a Quadratic Function

Ex: Find the zeros of the function, the line of symmetry, the vertex, the domain, the range and graph the parabola.

x

2 

6 x

 

Solution:

6

  

2 

4

Quadratic Formula x



2

6



32 x



2

6



4 2 x



2

The zeros of the function.

x 3 2 2





Zeros of a Quadratic Equation

Change the form of the function for easier graphing.

Add 0 9 9 inside the parenthesis.

y

 x

2 

6 x



1



 x

2 

6 x 9 9





1



 x

2 

6 x 9



9



1

  x



3



3 

8




Zeros of a Quadratic Function

Summary

Zeros:

Line of Symmetry:

Vertex:

 

Domain:





,



;

Range:

    

| x



3

 

8








Example 2: Zeros of a Quadratic Function

Find the zeros of the function, the line of symmetry, the vertex, the domain, the range and graph the parabola.

 

4 x

 

0

Solution:

Quadratic Formula.

x



4

  

2

2  

 

  

4



24 x





2

4



2 6 x





2

The zeros of the function.

x 2 6





Example 2: Zeros of a Quadratic Function (Cont.)

Change the form of the function.

Add 0 4 4 inside the parenthesis.

Put outside of the parenthesis.

y

  

4 x



2

 

 x

2 

4 x





2







 x

2 

4 x





 x

2 

4 x



4

4







4



4

 x 2



2 

6





2

2






Summary

Zeros: 2 6

Line of Symmetry:

Vertex:

Domain:



Range:



 

2,6









,





;

| x

 

2



6








Example 3: Zeros of a Quadratic Function

Find the zeros of the function, the line of symmetry, the vertex, the domain, the range and graph the parabola.

2 x

2 

6 x

 

0

Solution: x



6

  

2 

 

  

Quadratic Formula.

There are no real zeros

6

 

4 x

 because the discriminant

4 is negative. The graph will not cross the x-axis.





Example 2: Zeros of a Quadratic Function (Cont.) f x

  b

2 a



6



3

2 2

 

 



2

 

 

2



6

 

 

2



5

To find the vertex, use another approach. First find the x-value.

Plug the x-value into the original equation.

Simplify.

9 5



2

1

2



3 1

,





2 2



So we have the following vertex.






Summary

Zeros: No real zeros

Line of Symmetry:

Vertex:

Domain:





3 1

,





2 2





,



; x



3

2

Range: 



1

,



;  |

2







1

2



Insert graph p. 597 example

2 c.




Applications with Maximum/Minimum Values

Note:

The vertex of a vertical parabola is either the lowest point or the highest point on the graph of the parabola.

Therefore, the vertex can be used to determine the maximum or minimum value of a quadratic function.




Applications with Minimum/Maximum Values

Minimum and Maximum Values

For a parabola with its equation in the form y

   h



2  k ,

  is the lowest point and is called the minimum value of the function.

  maximum value of the function.




Application and Minimum/Maximum Values

If the function is in the general quadratic form y

 ax

2  bx

 c , then the maximum or minimum value x

 

2 b a y .

The concepts of minimum and maximum values of a function help not only in graphing but also in solving many types of applications.




Example 3: Maximization/Minimization Problems

A movie theater sells regular adult tickets for $7.50 each. On average, they sell 5,000 tickets per day. The company estimates that each time they raise ticket prices by 50¢, they will sell 1,000 fewer tickets. What price should they charge to maximize their revenue

(income) per day? What will be the maximum revenue?

Let x = number of 50¢ increases in price.

Then the price per ticket = 7.50 .50

x and the number of tickets sold =



,000 x .

Revenue = (price per unit)(number of units sold)


So,



Example 3: Maximization/Minimization

R

 

Problems (Cont.)

 x

  x



  x



2500 x

 

500 x

2 

5000 x



37,500



500 x 2

The revenue represented by a quadratic function and the maximum revenue occurs at the point where x

  b

2 a

 

2





5000



500



 

5 x

 

And the maximum revenue =







50

 

$50,000

000



$5

  




Example 4: Maximization/Minimization Problems

A farmer plans to use 2500 feet of spare fencing material to form a rectangular area for cows to graze alongside a river, using the river as one side of the rectangular area. How should he split up the fencing among the other three sides in order to maximize the rectangular area?

MOO!

x x

2500 2 x

If we let x represent the length of one side of the rectangular area then the dimensions of the rectangular area are x feet by 2500-2x feet (see image above). We will let A be the name of our function that we wish to maximize in this problem, so we want to find the

   




Example 4: Maximization/Minimization

Problems (Cont.)

Note: If we multiply out the formula for A , we get a quadratic function.

   x

  

2 x 2 

2500 x

We know this function is a parabola opening down. We also know that the vertex is the maximum point on this graph.

Remember, the vertex is the point







2 b 4 ac b

,

 a 4 a

2

 

 o r



 b

2 a

, f



 b

2 a 





.

So, plugging in the values, we get the vertex





625, A

  

.

Therefore, the maximum possible area is A(625):

A (625)



781250 square fee t .






Section 7.6a: Solving Quadratic Inequalities




Objectives o Solve quadratic inequalities.

o Solve higher degree inequalities.

o Graph the solutions for inequalities on real number lines.


Quadratic Inequalities



For values of x on either side of a number a, the sign for

 

  positive .

  negative . x a 0 any x

 a a x a 0 any x

 a




To Solve Quadratic or Higher Degree Inequalities

1. Arrange the terms so that one side of the inequality is 0.

2. Factor the algebraic expression, if possible, and find the points where each factor is 0. (Use the quadratic formula, if necessary, to find the points where a quadratic expression is 0.)

3. Mark these points on a number line. Consider these points as endpoints of intervals



Copyright © 2010 Hawkes Learning Systems. math courseware specialists

To Solve Quadratic or Higher Degree


Inequalities (Cont.)

4. Test one point from each interval to determine the sign of the expression for that interval.

5. The solution consists of those intervals where the test points satisfy the original inequality.




Example 1: Quadratic Inequalities

Solve the following inequality by factoring and using a number line. Then, graph the solution set on a number line. x

2 

2 x



15 x

2 

2 x



15



1 5



1 5 x

2 

2 x



15



0

 x



5

 x 3



0

Step 1: Add -15 to both sides so that the right-hand side is 0.

Step 2: Factor the left-hand side.




Example 1: Quadratic Inequalities (Cont.)

 x



5

 x 3



0 x 5 0 x

 

5 x 3 0 x



3

Step 3: Set each factor equal to

0 and solve to find the endpoints.

Step 4: Mark these points on a number line and test one point from each of the intervals.





5



3






Example 1: Quadratic Inequalities (Cont.)





7







5

 x



5

 x 3



0



Test



7

 x

 

7

5

 

7



3



 

5,3





0



3

  

10

 

20



0



3,

 



5

Step 5:

Test one point from each interval.

 x



5

 x 3



0 if x

 

5

Test x



0



0



5



0



3

    

15 0

 x



5

 x 3



0 if x 3

Test x



5



5



5



5



3

  

20



0

 x



5

 x 3



0 if x



3




Example 1: Quadratic Inequalities (Cont.) x

2 

2 x



15

We have found that:

 x



5

 x 3



0 if x

 x



5

 x 3



0 if

 x



5

 x 3



0 if x

 

5



3 x 3

Therefore, the solution is: x

 

5 or or x is in

 x



3

, 5

 

3,

 

.

)



5 3

(




Example 2: Cubic Inequality

Solve the following inequality by using a number line and graph each solution set on a number line. x

3 

8 x

2 

9 x



0



2 

8 x



9





0

 

1

 x 9



0

Factor the left-hand side. x



0 x



0 x 1 0 x

 

1 x 9 0 x



9

Set each factor equal to 0 and solve to find the endpoints.




Example 2: Cubic Inequality (Cont.)

Mark the endpoints on a number line and test one point from each of the intervals found.

Endpoints : x

 

1

 





2



1

 

1, 0



0



1

 

1

 x 9



0

2

Test x

 

2 x



0

 



3

  

2



1

 

2



9

 x



9





9,







9 10

   

11



22 0

Test x



3

  

1



3



9

 

72 0




Example 2: Cubic Inequality (Cont.)

 

1

 x



9

 

0



2

 



1



1

0

2

Test x

 

1

2





1

2





1

2



1



1

2



9



Test x



10

 

10



1



10



9

 



3 9

 

10



1

2

1

2



19

2

 

19



8

0



110



0




Example 2: Cubic Inequality (Cont.) x

3 

8 x

2 

9 x



0

We have found that:

 

1

 x



9

 

0 if x

 

1

 x



9

 

0 if 0

 

1

 

1

 x



9

 

0 if x



9

9

 

1

 x



9

 

0 if x 0

Therefore, the solution is x is in



, 1

  



1

 

0



9




Example 3: Quadratic Inequality

Solve the following inequality by using a number line and graph each solution set on a number line. x

2 

6 x

 

0

Solution: Since this quadratic equation will not factor with integer coefficients, use the quadratic formula to find the endpoints for the intervals. The test points can be integers. x



  b

2 

4 ac

2 a x



6

  

2 

 

   x



6



36 28

2




Example 3: Quadratic Inequality (Cont.) x



6



36 28

2 x



6



8

2 x



2 x 3 2

0



3



2



3



2




Example 3: Quadratic Inequality (Cont.)

 x

2 

6 x

 

0

  

0 3



2

Test x



0

 

2 

6

  

7 0 0 7 7 0

3 3



2

Test x



3

 

2 

6

  

7 9 18 7 2 0

Test x



7

 

2 

6

  

7



49 42 7



14



0



7




Example 3: Quadratic Inequality (Cont.) x

2 

6 x

 

0

We have

If x 3 2 then x 2 

6 x 7 0

If x 3 2 then x 2 

6 x 7 0

The solution is: x is in



,3 2

 

3 2,





.



3



2



3



2






Section 7.6b: Solving Inequalities with

Rational Expressions




Objectives o Solve inequalities containing rational expressions.

o Graph the solutions for inequalities containing rational expressions.




Solving Inequalities with Rational Expressions o For linear inequalities, the solution always contains the points to one side or the other of the point where the inequality has the value 0. o A rational inequality may involve the product or quotient of several first-degree expressions. For example, x x



4



3



0 o For ( ), if , then is negative ; if , is positive .




Solving Inequalities with Rational Expressions

Procedure for Solving Polynomial Inequalities with

Rational Expressions

1. Simplify the inequality so that one side is 0 and the other side has a single fraction with both the numerator and denominator in factored form.

2. Find the points that cause the factors in the numerator or in the denominator to be 0.

3. Mark each of these points on a number line.

These are interval endpoints.






4. Test one point from each interval to determine the sign of the polynomial expression for all points in that interval.

5. The solution consists of those intervals where the test points satisfy the original inequality.

6. Mark a bracket for an endpoint that is included and a parenthesis for an endpoint that is not included. Remember that no denominator can be 0.





The steps are as follows: x x





4



0

3

a. Find the points where each linear factor has the v value 0. x 4 0 x 3 0 x

 

4 x



3

b. Mark each of these points on a number line.

C Consider these points as endpoints of intervals.





4 0

The three intervals formed are



3

      

.





c. Choose any number from each interval as a test value to determine the sign of the expression for all values in that interval. Remember, we are not interested in the value of the expression, only whether it is positive or negative.

 



6

 



4

 

4,3



0





3





4

3,

 

4







6 0 gives: x



4 x



3 x



4 x



3 x



4 x



3







Results

 

 

4

3







2

9



 

 





4

3





4

3

 

 





4

3



8

1





 

8

2

9

4

3



4



Explanation

The expression is positive for all x in

 

The expression is negative for all x in



(-4,3)




3










Solving Inequalities with Rational Expressions x x



4



3



0

d. The solution to the inequality consists of all the intervals that indicate the desired sign: + (for ) or

– (for ). The solution for the above expression is: all in



, 4



or



3,

 

.

In interval notation: x

 

4



, 4

 

3.

3,

 

.

Graphically:



4

 

3




Example 1: Solving and Graphing Inequalities

Solve and graph the solution for the following inequality: x x





1

5



0

Solution: Set each linear expression equal to 0 to find the interval endpoints. x 1 0 x

 

1 x 5 0 x



5

Test a value from each of the intervals:

     

1,5



, and



5,

 





3





1



0



5



6





Example 1: Solving and Graphing




3 0 gives: x



1 x



5

 x x



1



5

 x x



1



5



Results

 

  

 

 



1

0



5

1

5





1



5



2



8



 

 

 







1

5



7

7

1





1

1

4

1

5

Explanation


 




(-1,5).






 



5






Inequalities (Cont.) x x



1



5



0

We can conclude that the solution for the above inequality in interval notation is:

  

In algebraic notation: 1 x 5.

Graphically:

(



1

)

5




Example 2: Solving and Graphing Inequalities

Solve and graph the solution for the following inequality: x



6 x



2

Solution: Simplify to get one fraction. x

 x



6

2

 x x



2



2



0

2 x



4 x



2



0







Set each linear expression equal to 0 to find the interval endpoints.

2 x 4 0 x

 

2 x 2 x



0

2

Test a value from each of the intervals:

     

2, 2



, and



2,

 



4

  



2 0



2



4







Substituting the values -4, 0, and 4 into the original inequality gives:

2 x



4 x



2



2 x



4 x



2



2 x



4 x



2



Results

2

 

  

2

4

 

 



0



2

4





4



4



6



2



 

2

2

3

 

 







2

4



2



12

2



6



Explanation


 




(-2,2).




 



2






Inequalities (Cont.) x

 x



6

2

We can conclude that the solution for the above inequality in interval notation is:



, 2

 

2,

 

.

In algebraic notation: x

 

2 or x



2.

Graphically:





2 2

(

Note: 2 from the denominator is not included in the solution set because the quotient is undefined there.

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